How many lone pairs are present in `OF_2` molecule ?

Explanation

To determine the number of lone pairs present in the `OF_2` molecule, we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Oxygen (O)** has an atomic number of 8, and its electron configuration is `1s² 2s² 2p⁴`. Therefore, oxygen has 6 valence electrons (2 from 2s and 4 from 2p).
- **Fluorine (F)** has an atomic number of 9, and its electron configuration is `1s² 2s² 2p⁵`. Thus, each fluorine atom has 7 valence electrons.

### Step 2: Calculate Total Valence Electrons in `OF_2`
- In `OF_2`, there is 1 oxygen atom and 2 fluorine atoms.
- Total valence electrons = (Valence electrons of O) + 2 * (Valence electrons of F)
- Total valence electrons = 6 + 2 * 7 = 6 + 14 = 20 valence electrons.

### Step 3: Draw the Lewis Structure
- Place the oxygen atom in the center because it is less electronegative than fluorine.
- Connect each fluorine atom to the oxygen atom with a single bond. Each bond uses 2 electrons.
- After forming 2 bonds (1 bond to each fluorine), we have used 4 electrons (2 for each bond).
- Remaining electrons = 20 - 4 = 16 electrons.

### Step 4: Distribute Remaining Electrons
- Each fluorine atom needs 6 more electrons to complete its octet (since they already have 2 from the bond).
- Place 6 electrons (3 lone pairs) around each fluorine atom.
- Now we have used 12 electrons (6 for each fluorine), leaving us with 4 electrons.

### Step 5: Assign Remaining Electrons to Oxygen
- Place the remaining 4 electrons around the oxygen atom. These will form 2 lone pairs on the oxygen atom.

### Step 6: Count the Lone Pairs
- The oxygen atom has 2 lone pairs.
- Each fluorine atom has 3 lone pairs.
- Total lone pairs = 2 (from O) + 3 (from F1) + 3 (from F2) = 8 lone pairs.

### Final Answer
The `OF_2` molecule has a total of **8 lone pairs**.

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